Chemistry Assessments
Chem 16A
- a) 2HCl(aq) + Mg(OH)2(aq) → MgCl2(aq) + 2H2O(l)
b) Using 22ml of 0.7M HCl and 14.7ml of 0.55M Mg(OH)2 the reaction would proceed as follows;
The mole ratio for the reactants is 2:1 for each mole of Mg(OH)2, two moles of HCl are required. Thus, for 0.008085Moles of Mg(OH)2 to be fully consumed, there must be 0.01617 moles of HCl. This is less than the 0.0154 moles provided making HCl the limiting reagent (Da Silva, 2017). Thus, the products formed will depend on the moles of HCl used.
The mole ration of HCl to MgCl2 is 2:1. Therefore, two moles of HCl produce 1 mole of MgCl2. Since 0.0154 moles of HCl were used, then of MgCl2 were produced. Since the molar mass of MgCl2 is 95.211g/mol, it follows that the mass is
- a) H2SO4(aq) + Ca(OH)2(aq) → CaSO4(aq) + 2H2O(l)
b) Using 36 ml of 0.2M H2SO4(aq) and 20 ml of 0.1M Ca(OH)2(aq);
Since the mole ratio of the reactants is 1:1, then Ca(OH)2 is the limiting reagents since it will be depleted first (Da Silva, 2017). It will therefore determine the moles of H2O produced as follows;
Mole ratio between Ca(OH)2 and water is 1:2, it follows that using 0.002 moles of Ca(OH)2, 0.004moles of water are produced. The molar mass of water is 18.0153g/mol and therefore, mass of water produced is
c).
- HBrO3(aq) + 5HBr(aq) → 3Br2(g) + 3H2O(l)
- Using 100ml of 2M HBrO3(aq) and 10g of HBr, one gets;
Since the mole ratio of reactants is 1:5, is required to consume the 0.2 moles of HBrO3. Since the available moles of HBr are 0.1236 less than the required, it is the limiting reagent. The moles of HBr produced are therefore dependent on moles of HBr used. The mole ratio of HBr to Br2 is 5:3, thus, for every mole of HBr consumed, 3/5moles of Br2 are produced. Since 0.1236moles were used, it follows that The molar mass of Br2 is 159.808g/mol. Theoretical yield is 159.808g/mol×0.07416mol=11.8514g
- 2Al + 3Pb(NO3)2 → 3Pb + 2Al(NO3)3
Using 15g of Al and 450ml of 2M 3Pb(NO3)2;
The mole ratio of reactants is 2:3 or 1:1.5 meaning 0.556moles of Al will require 0.556×1.5 = 0.834moles of Lead (II) Nitrate. This is within the moles provided making Al the limiting reagent. As such, the moles of lead (the solid) are;
Mole ratio 2:3 or 1:1.5, 0.556 moles of Al were consumed which means 0.834moles of Pb were produced. Molar mass of Pb is 207.2g/mol. Thus, Mass=207.2g/mol×0.834=172.8048g.
- 2Al(NO3)3(aq) + 3Ca(OH)2(aq) → 3Ca(NO3)2(aq)+ 2Al(OH)3(s)
500ml of 0.15M Al(NO3)3 and 400ml of 0.2M Ca(OH)2, the yield is as follows;
The mole ratio of the reactants is 2:3 or 1:1.5 which means 0.075 moles of Aluminum Nitrate will require 0.075×1.5 moles=0.1125 moles of calcium hydroxide. This is more than the available amount making calcium hydroxide the limiting reagent. Mole ratio between calcium hydroxide and aluminum hydroxide is 3:2 or 1.5:1. Thus, 0.08 moles will produce Its molar mass is 78g/mol and therefore mass= 78g/mol×0.05333mol=4.16g
Chem 16B
- Magnesium reacts with hydrochloric acid to form magnesium chloride and hydrogen gas. It is the reducing agent because it loses electrons to hydrogen ions. On the other hand, Hydrochloric acid is the source of hydrogen ions (H+) and oxidizing agent because it accepts the electrons leading to the formation of hydrogen gas. In the first four balloons, Magnesium is the limiting reagent. This is because it is depleted first in the reaction (Da Silva, 2017). HCl is in excess because it can be observed that as the amount and, therefore, moles of Mg increase, the amount of gas released increases judging by the size of the balloons about a minute later (Jim Holler, 2012). This means the volume of the product (hydrogen) depends on the amount of magnesium.
In the last three balloons, some magnesium remains unconsumed, and it can be observed that the size of the balloons is comparatively similar. This is because all the HCl has been consumed, and there is none left to react with the remaining magnesium. This makes HCl the limiting reagent and Mg the excess reagent because HCl hindered Hydrogen production.
- Unbalanced Equation; Mg(s) + HCl(aq) →H2(g) + MgCl2(aq)
Balanced Equation; Mg(s) + 2HCl(aq) →H2(g) + MgCl2(aq)
From the video, it can be observed that 82mmol of HCl(aq) reacted with 41mmols of Magnesium in the balloon where the reactants were all consumed (Jim Holler, 2012). This makes it a 2:1 mole ratio used to balance the above equation.
Chem 17A
- Sulfur dioxide undergoes oxidation in the atmosphere in the presence of atmospheric oxidants such as ozone, hydroxyl radicals, and oxygen forming sulfur trioxide. Similarly, nitric oxide (NO) reacts with the atmospheric oxidants to form Nitrogen dioxide. The following equation represents the chemical changes; SO2(g) + O3 (g)→ SO3 (g)+ O2 (g) and NO(g) + O3 (g)→ NO2 (g)+ O2(g). In the case of sulfur trioxide, hydration occurs through the following equation SO3(g) + H2O(l) → H2SO4(aq), forming sulfuric acid, which is a strong acid that, when absorbed in the atmosphere and is part of the humidity forming clouds and therefore the rain (GitHub, 2023). Similarly, nitrogen dioxide undergoes hydration to form nitric acid through the following reaction 3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g) (GitHub, 2023). Once formed, nitric acid can dissolve in atmospheric moisture, such as clouds and fog droplets, forming aqueous nitric acid. This nitric acid contributes to the acidity of the precipitation. When absorbed in water, it dissociates.
- 3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g).
Half equations; NO2 → HNO3 In this half equation, Nitrogen is oxidized from +4 to +5.
H2O + 2e- → H2 + 2OH-. Here, water is reduced by gaining two electrons, forming hydrogen and two hydroxyl ions. Combining the two half equations, one gets; 3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g), where Nitrogen is oxidized, and water is reduced.
- Reducing climate change and the formation of acid rain by minimizing nitrogen oxides and sulfur dioxide is a moral obligation. The following is my action plan to cut down on these emissions. First, I will reduce fossil fuel use. Fossil fuels majorly contribute to nitrogen oxides, especially coal (EPA, 2023). By minimizing energy use at home and boarding buses and trains whenever possible, I will cut down on the amount of fossil fuels used and hence nitrogen oxides and sulfur dioxide. Further, I can ensure that family vehicles are using energy efficiently and that there is proper filtration of the emissions by regularly servicing the vehicles. Second, as a family, we can minimize energy use by recycling the available energy; for example, by insulating my room and the rest of the house, I do not need to warm it, reducing the amount of energy used. Third, I can recycle plastics and properly dispose of garbage to enhance recycling and reduce the manufacturing of these products. Fourth, I can be an environmental activist informing those around me on how to reduce emissions of these gasses through sustainable agriculture, energy saving, and requiring environmentally friendly practices by their firms.
17b
Dissociation and Concentration
Sodium Hydroxide is a strong base and therefore dissociates fully (100%) to produce sodium and hydroxyl ions with a concentration of 1M each since 1M NaOH was used. This is represented as follows.
Before; NaOH → Na+ + OH-
100% or 1M 0
After; NaOH → Na+ + OH-
0 100% or 1M
In contrast, ammonium hydroxide is a weak base that undergoes partial dissociation in water to form ammonium ions and hydroxyl ions. Represented in an equation, the dissociation will be as follows.
Before; NH4OH ⇌ NH4+ + OH-
100% or 1M 0
After; NH4OH ⇌ NH4+ + OH-
99% or 0.99M 1% or 0.01M
Conductivity and pH
An increase in the concentration of ions in a solution is associated with an increase in conductivity (Atlas Scientific, 2022). As such, a solution with more free ions will have higher conductivity. In this case, sodium hydroxide has 1M free ions compared to ammonium hydroxide, which has 0.01M. It is expected that sodium will have a significantly higher conductivity.
With regard to pH, pH is a measure of the relative amount of free hydrogen and hydroxyl ions, which means that water with more hydrogen or hydroxyl ions is more acidic or basic, respectively. In this case, sodium hydroxide has more free hydroxyl ions, 1M, than ammonium hydroxide, 0.01M. Its pH would be, therefore, higher, closer to 14, than ammonium hydroxide’s, which would be closer to 7.
Reactivity
The reactivity of a base is determined by the concentration of hydroxyl ions. An increase in concentrations means there are more collisions among reacting molecules and, therefore, a higher rate of reaction. As such, since sodium hydroxide has a greater concentration of hydroxyl ions, one can expect that more collisions among reacting species will increase the reaction rate compared to ammonium hydroxide. This is the collision theory.
Chem 18a
- 2HCl(aq) + Ca(OH)2(aq) → CaCl2(aq) + 2H2O(l)
20ml sample of HCl required 16 ml of 2M calcium hydroxide.
Since the mole ratio is 2:1, and the moles of calcium hydroxide used were 0.032, it follows that moles of HCl in the sample were 0.032×2=0.064 moles. The concentration of the sample was therefore
- 2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + H2O(l) + CO2(g)
Using a 15ml sample that was neutralized by 10ml of 0.35M Na2CO3(aq).
The mole ratio is 2:1, and 0.0035moles were consumed, as such, the moles of HCl in the sample were; 0.0035×2=0.007moles. The concentration is therefor,
- HF(aq) + KOH(aq) → KF(aq) + H2O(l)
Waste drum has 90L of 5.6M HF. Finding the moles of HF;
Since the mole ratio of the reactants is 1:1 and the moles of HF in the drum are 504, then the moles of KOH required to neutralize the drum are 504 moles.
- H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
10ml water sample was neutralized by 9.56ml of 0.0412M NaOH.
Mole ratio is 1:2 meaning the moles of sulfuric acid is . Thus, the concentration is;
Chem 18b
An analysis of the table shows the pH of the water is at 8 in point A, 4.3 at point B, and then rises to 6.0 at point C. The ideal water pH is between 6.5 and 8.2 (Water Quality Tests Summary, n.d.). pH below 6 increases the amount of mercury soluble in water and leads to physical damage to the gills, exoskeleton, and fins of the fish. Given the trend in the chart, the pH is lowered by substances released by the metal stripping plant. Heavy metals released by the plant are more toxic due to pH.
The NO3- and PO43- are constant at points A and B at 0.2 ppm and increase at point C to 5 ppm and 6 ppm, respectively. Animal feedlots, stockyards, and agricultural runoff are considered major contributors to nitrate ions in water bodies (Water Quality Tests Summary, n.d.). The pig farm is, therefore, the most likely source of the nitrates. However, nitrates are toxic at close to 10 ppm and not 5 ppm (Water Quality Tests Summary, n.d.). They are an unlikely cause of fish death. The source of phosphates is likely agricultural waste from the pig farm. Since the phosphates are significantly higher than the recommended 0.1ppm, there might be eutrophication, algae bloom, and decreased dissolved oxygen that might kill fish.
The dissolved oxygen decreases from point A, 8 ppm, to point B, 7 ppm, and more on point C, 3 ppm. The increase in Nitrates can explain this, and phosphates in the river are due to agricultural and livestock wastes from the pig farm. A minimum of 5 ppm of dissolved oxygen is required for aquatic life. The suspended solids are low in point A, high in point B, and medium in point C. This is associated with the acidity of the water that favors the accumulation of metallic waste from the metal stripping plant.
Thus, both the pig farm and metal stripping plant are contributing to fish death. The metal stripping plant acidifies water and releases metallic waste favoring their accumulation. This is toxic to the fish and leads to death. The pig farm releases phosphates and nitrates, reducing the dissolved oxygen, which deprives aquatic life of oxygen, leading to death.